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Question
Prove the following identity :
`tan^2A - tan^2B = (sin^2A - sin^2B)/(cos^2Acos^2B)`
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Solution
LHS = `tan^2A - tan^2B`
= `sin^2A/cos^2A - sin^2B/cos^2B`
= `(sin^2Acos^2B - cos^2Asin^2B)/(cos^2Acos^2B)`
= `((1 - cos^2A)cos^2B - cos^2A(1 - cos^2B))/(cos^2Acos^2B)`
= `(cos^2B - cos^2Acos^2B - cos^2A + cos^2Acos^2B)/(cos^2Acos^2B)`
= `(cos^2B - cos^2A)/(cos^2Acos^2B)`
= `((1 - sin^2B) - (1 - sin^2A))/(cos^2Acos^2B)`
= `(sin^2A - sin^2B)/(cos^2Acos^2B)`
Hence `tan^2A - tan^2B = (cos^2B - cos^2A)/(cos^2A cos^2B) = (sin^2A - sin^2B)/(cos^2Acos^2B)`
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